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Comment by vl | Hacker News Reader

Not even close!

An easy way to make upper boundary estimate: let’s say you use Falcon Heavy to deliver your kinetic weapon. Falcon Heavy burns around 500 tons of fuel. This is upper boundary for energy of your payload (since most of the energy goes to lifting rocket itself and air resistance). So even ultra-heavy payload of 64 tons will have way less then 0.5 kilotons of energy (TNT equivalent used for nukes). Any nuke is way beyond that.

Of course, another way to estimate is to not to be lazy and actually multiply mv^2/2. Truly massive 64 ton kinetic weapon reaching target at 7 km/s will have energy of 1.6x10^12 J. Modest 20 kiloton nuke has energy of 8.4x10^13 J. I.e. 50 times more.

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> let’s say you use Falcon Heavy to deliver your kinetic weapon.

Why? pick up an NEO and aim for Sydney, or wherever it needs to go. Easy, unless your settlement isn't orbital or you don't have electric propulsion.

What is a NEO?

Near-earth orbit

Thanks. :)

Near-Earth Object. Asteroids around Earth.

There is infinite supply of all sizes of boulders in space that you can just grab and turn into missiles given you have sufficient level of technology.

Thank you. Is there a sci-fi book where NEOs are used that you'd recommend? I'm not too versed in that genre.

(Roughly speaking) Kinetic weapon has as much energy as you spent getting it on orbit from surface, thus fuel in the rocket provides useful estimate and comparison point for energy available. To get from orbit to surface you actually need to slow down, ie reduce energy.

I'm saying you can pick up a large enough rock floating in solar orbit and nudge it to coincide with Earth using electric propulsion. Basically forcing a dinosaur wipeout event in a small scale. Sky's the limit for energy available.

But the problem is that for any rock large enough it will require a lot of energy and time, thus expensive and gives advanced warning.

You're assuming the kinetic weapon originates on Earth, which doesn't apply when talking about a weapon wielded by a space colony against Earth. Instead, you should be calculating how much rock or metal a Falcon Heavy could lift off the surface of the Moon and set onto an Earth-intercepting trajectory, then subtract the kinetic energy that will be lost during reentry (hence the suggestion to use tungsten).

No, I used this estimate as simple way to calculate TNT equivalent. Of course energy of collision is proportional to delta-v. In solar system it’s tricky (or rather very energy-demanding) to achieve delta-v with Earth exceeding 30km/s. Which would require truly massive projectile to compare in energy with nuke.

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